Choose the correct option:
A wire of length `l` and mass `m` is first bent in a circle, then in a square and then in an equilateral triangle. The moment of inertia in these three cases about an axis perpendicular to their planes and passing through their centre of mass are `I_(1),I_(2)` and `I_(3)` respectively. Then maximum of them is

Options

A. (a) `I_(1)`
B. (b) `I_(2)`
C. (c) `I_(3)`
D. (d) Data insufficient

Answer

To solve the problem, we need to calculate the moment of inertia for a wire bent into three different shapes: a circle, a square, and an equilateral triangle. We will denote the moments of inertia for these shapes as \( I_1 \), \( I_2 \), and \( I_3 \) respectively.

### Step 1: Calculate the moment of inertia for the circle

1. **Circumference of the circle**: The circumference \( C \) is equal to the length of the wire \( L \):
\[ C = 2\pi r \implies L = 2\pi r \implies r = \frac{L}{2\pi} \]

2. **Moment of inertia for a circular wire**: The moment of inertia \( I_1 \) about an axis perpendicular to the plane of the circle and through its center of mass is given by:
\[ I_1 = m r^2 = m \left(\frac{L}{2\pi}\right)^2 = \frac{mL^2}{4\pi^2} \]

### Step 2: Calculate the moment of inertia for the square

1. **Perimeter of the square**: The perimeter \( P \) is equal to the length of the wire \( L \):
\[ P = 4a \implies L = 4a \implies a = \frac{L}{4} \]
where \( a \) is the length of one side of the square.

2. **Moment of inertia for a square wire**: The moment of inertia \( I_2 \) about an axis perpendicular to the plane of the square and through its center of mass is given by:
\[ I_2 = \frac{1}{6} m (2a^2) = \frac{1}{6} m (2 \cdot \left(\frac{L}{4}\right)^2) = \frac{mL^2}{96} \]

### Step 3: Calculate the moment of inertia for the equilateral triangle

1. **Perimeter of the equilateral triangle**: The perimeter \( P \) is equal to the length of the wire \( L \):
\[ P = 3b \implies L = 3b \implies b = \frac{L}{3} \]
where \( b \) is the length of one side of the triangle.

2. **Moment of inertia for an equilateral triangle wire**: The moment of inertia \( I_3 \) about an axis perpendicular to the plane of the triangle and through its center of mass is given by:
\[ I_3 = \frac{1}{6} m b^2 = \frac{1}{6} m \left(\frac{L}{3}\right)^2 = \frac{mL^2}{54} \]

### Step 4: Compare the moments of inertia

Now we have the expressions for the moments of inertia:
- \( I_1 = \frac{mL^2}{4\pi^2} \)
- \( I_2 = \frac{mL^2}{96} \)
- \( I_3 = \frac{mL^2}{54} \)

To determine which is the maximum, we can compare the denominators:
- \( 4\pi^2 \approx 39.478 \)
- \( 96 \)
- \( 54 \)

Since \( 4\pi^2 < 54 < 96 \), we find that:
\[ I_1 > I_3 > I_2 \]

### Conclusion

Thus, the maximum moment of inertia is:
\[ \text{Maximum of } (I_1, I_2, I_3) = I_1 \]

The correct option is **A, \( I_1 \)**.